TSTP Solution File: SEU843^5 by cocATP---0.2.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEU843^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n180.star.cs.uiowa.edu
% Model    : x86_64 x86_64
% CPU      : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz
% Memory   : 32286.75MB
% OS       : Linux 2.6.32-431.20.3.el6.x86_64
% CPULimit : 300s
% DateTime : Thu Jul 17 13:33:16 EDT 2014

% Result   : Theorem 0.53s
% Output   : Proof 0.53s
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----ERROR: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SEU843^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n180.star.cs.uiowa.edu
% % Model    : x86_64 x86_64
% % CPU      : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
% % Memory   : 32286.75MB
% % OS       : Linux 2.6.32-431.20.3.el6.x86_64
% % CPULimit : 300
% % DateTime : Thu Jul 17 11:36:11 CDT 2014
% % CPUTime  : 0.53 
% Python 2.7.5
% Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/']
% FOF formula (<kernel.Constant object at 0x1ba77e8>, <kernel.Type object at 0x1ba71b8>) of role type named a_type
% Using role type
% Declaring a:Type
% FOF formula (forall (S:(a->Prop)) (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx))))) of role conjecture named cGAZING_THM23_pme
% Conjecture to prove = (forall (S:(a->Prop)) (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx))))):Prop
% Parameter a_DUMMY:a.
% We need to prove ['(forall (S:(a->Prop)) (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx)))))']
% Parameter a:Type.
% Trying to prove (forall (S:(a->Prop)) (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx)))))
% Found or_intror00:=(or_intror0 (T Xx)):((T Xx)->((or (S Xx)) (T Xx)))
% Found (or_intror0 (T Xx)) as proof of ((T Xx)->((or (S Xx)) (T Xx)))
% Found ((or_intror (S Xx)) (T Xx)) as proof of ((T Xx)->((or (S Xx)) (T Xx)))
% Found (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))) as proof of ((T Xx)->((or (S Xx)) (T Xx)))
% Found (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))) as proof of (forall (Xx:a), ((T Xx)->((or (S Xx)) (T Xx))))
% Found (x0 (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx)))) as proof of (forall (Xx:a), ((T Xx)->(S Xx)))
% Found ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx)))) as proof of (forall (Xx:a), ((T Xx)->(S Xx)))
% Found (fun (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))))) as proof of (forall (Xx:a), ((T Xx)->(S Xx)))
% Found (fun (T:(a->Prop)) (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))))) as proof of ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx))))
% Found (fun (S:(a->Prop)) (T:(a->Prop)) (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))))) as proof of (forall (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx)))))
% Found (fun (S:(a->Prop)) (T:(a->Prop)) (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx))))) as proof of (forall (S:(a->Prop)) (T:(a->Prop)), ((((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S)->(forall (Xx:a), ((T Xx)->(S Xx)))))
% Got proof (fun (S:(a->Prop)) (T:(a->Prop)) (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx)))))
% Time elapsed = 0.220874s
% node=20 cost=-130.000000 depth=9
% ::::::::::::::::::::::
% % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% (fun (S:(a->Prop)) (T:(a->Prop)) (x:(((eq (a->Prop)) (fun (Xz:a)=> ((or (S Xz)) (T Xz)))) S))=> ((x (fun (x1:(a->Prop))=> (forall (Xx:a), ((T Xx)->(x1 Xx))))) (fun (Xx:a)=> ((or_intror (S Xx)) (T Xx)))))
% % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
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